Calculating conditional expectation

Question

I have to calculate $E\left(\int_1^4 W_t^3dt |\mathcal F_2\right)$

My solution:

$E(\int_1^4 W_t^3dt |F_2)=E(\int_1^2 W_t^3dt |F_2)+E(\int_2^4 W_t^3dt |F_2)=\int_1^2 W_t^3dt+\int_2^4 E(W_t^3 |F_2)dt$

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$E(W_t^3|F_2)=E((W_t-W_2+W_2)^3|F_2)=E((W_t-W_2)^3|F_2)+3E((W_t-W_2)^2 W_2|F_2)+3E((W_t-W_2) W_2^2|F_2)+E(W_2^3|F_2)=3W_2 (t-2)+W_2^3$

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So $\int_2^4 E(W_t^3 |F_2)dt=6W_2+2W_2^3$

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So our result is: $\int_1^2 W_t^3dt+6W_2+2W_2^3$

Is that ok?

Answer 1

It is not specified in the question, but I assume that $(W_t)_{t\geqslant 0}$ is a standard Brownian motion and that $\mathcal F_2$ is the $\sigma$-algebra generated by the $W_t, t\leqslant 2$.

The computations rest on the independence of $W_t-W_2$ and $\mathcal F_2$ for $t\gt 2$ and that $\int_1^2W_s\mathrm ds$ is measurable with respect to $\mathcal F_2$. The latter need to be detailed. The computations appear to be correct.